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Is there some precise sense in which the "alternation" functor $A$ that maps a multilinear function $f\colon M^d\to N$ to the alternating multilinear function $A(f)\colon M^d\to N$ defined by $$ A(f)(x_1,\dotsc,x_d) = \sum_{\sigma\in S_n}\operatorname{sign}(\sigma) f(x_{\sigma(1)},\dotsc,x_{\sigma(d)}) $$ is the canonical or most natural one? Here $M$ and $N$ are vector spaces or modules over a commutative ring $R$.

In particular, is there a mathematical reason, when one needs to "alternate" a multilinear function, to choose this $A$ over $-A$?

Note. I did not use the factor $\frac{1}{d!}$ because in a ring or a field $d!$ does not need to have an inverse.

The same question can be asked i suppose about the "symmetrization" functor $S$ defined by $$ S(f)(x_1,\dotsc,x_d) = \sum_{\sigma\in S_n} f(x_{\sigma(1)},\dotsc,x_{\sigma(d)}). $$

A related or maybe equivalent question: in which sense the maps $$ M^{\otimes d}\to M^{\otimes d},\quad x_1\otimes\dotsb\otimes x_d\mapsto \sum_{\sigma\in S_n}\operatorname{sign}(\sigma) x_{\sigma(1)}\otimes\dotsb\otimes x_{\sigma(d)} $$ and $$ M^{\otimes d}\to M^{\otimes d},\quad x_1\otimes\dotsb\otimes x_d\mapsto \sum_{\sigma\in S_n} x_{\sigma(1)}\otimes\dotsb\otimes x_{\sigma(d)} $$ are "canonical"?

I think "canonical" should mean "determined uniquely up to isomorphism by some universal property."


Sometimes signs in standard definitions are chosen somewhat or completely randomly.

Example. In the axiomatic definition of the exterior derivative of a differential form, there is absolutely no reason to require $$ \mathrm{d}(α ∧ β) = \mathrm{d}α ∧ β + (−1)^p (α ∧ \mathrm{d}β), $$ where $α$ is a $p$-form. It could be as well required instead that $$ \mathrm{d}(α ∧ β) = (−1)^q (\mathrm{d}α ∧ β) + α ∧ \mathrm{d}β, $$ where $β$ is a $q$-form. (It seems to me that the most general form of a differential algebra product rule should be something like this: $d(ab) = d(a)\mu(b) + \lambda(a)d(b)$, where $\lambda$ and $\mu$ are two fixed endomorphisms.)


Note about terminology. I must admit that the terms like "symmetrization" (in the title, etc.) are probably misleading and badly chosen, because it is usually assumed that symmetrizing symmetric things should not change them.

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    $\begingroup$ You could also ask what makes the mapping $V \rightarrow V^{**}$ by $v \mapsto (\varphi \mapsto \varphi(v))$ from vector spaces to their double-dual spaces the "most natural" one. For a fixed nonzero $c$ in the underlying scalar field, the mapping $V \rightarrow V^{**}$ by $v \mapsto (\varphi \mapsto c\varphi(v))$ could serve a similar purpose. Why do you consider the choice $c = 1$ to be best? $\endgroup$
    – KCd
    Jun 7, 2014 at 9:39
  • $\begingroup$ When $M$ is a free $R$-module, the mappings you write down at the end of your question lead to embeddings of the $d$th exterior and symmetric powers of $M$ into the $d$th tensor power of $M$. This can be used to prove exterior and symmetric powers of free modules over any commutative ring are free (when they're not 0). That is a useful application of these mappings. See Theorem 4.2 of math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf for the case of exterior powers. (Corollary 5.10 is another nice use of this mapping for free modules.) $\endgroup$
    – KCd
    Jun 7, 2014 at 9:45
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    $\begingroup$ @KCd: one way in which i consider $v\mapsto(\phi\mapsto\phi(v))$ more natural than, for example, $v\mapsto(\phi\mapsto-\phi(v))$ is that it takes one operation less to define. I will think if i can give a more formal reason to call it "canonical". $\endgroup$
    – Alexey
    Jun 7, 2014 at 10:09
  • $\begingroup$ Small note: The reason for taking the "usual" definition of $A$ rather than (say) $-A$ is presumably so the "alternation map" will be the identity on alternating tensors. (This point seems not to have been made yet, but admittedly I haven't read each answer and comment carefully.) $\endgroup$ Apr 3, 2015 at 12:42
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    $\begingroup$ Good point, @user86418, but then a coefficient $1/d!$ is also needed. $\endgroup$
    – Alexey
    Apr 3, 2015 at 12:53

2 Answers 2

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I claim that the "most natural" option involves those factors of $\frac{1}{d!}$ that you omitted. The reason is the following. $M^{\otimes d}$ is naturally equipped with a representation of the symmetric group $S_d$. At least if we're working over a field, it's natural to try to isolate the isotypic component of the trivial resp. the sign representation (the symmetric resp. the antisymmetric tensors), and if we're working in characteristic zero, it's natural to do this using the canonical idempotents in the group algebra $\mathbb{Q}[S_d]$ that do this, namely

$$\frac{1}{d!} \sum_{\pi \in S_d} \pi$$

and

$$\frac{1}{d!} \sum_{\pi \in S_d} \text{sgn}(\pi) \pi$$

respectively. The point of those divisions by $d!$ is so that the resulting operations are idempotents, and in particular so that they fix the subspace of symmetric resp. antisymmetric tensors. More generally, if a finite group $G$ acts on a vector space $W$ and $V$ is an irreducible representation of $G$, then the idempotent $\frac{1}{|G|} \sum_{g \in G} \overline{\chi_V}(g) g$ is the canonical idempotent that always projects onto the $V$-isotypic component, where $\chi_V$ is the character of $V$.

The fact that it's natural to divide by $d!$ in this argument suggests that weird things happen when you can't, and indeed they do. If you're working in characteristic $p$ where $p \le d$ then you enter the realm of modular representation theory, where Maschke's theorem fails in general. In particular, there's no reason to expect the symmetric tensors to be a direct summand of $M^{\otimes d}$ as an $S_d$-representation in general, so there is no reason to expect that there exists an idempotent commuting with the action of $S_d$ that projects onto it. If you try to symmetrize without dividing by $d!$ then your symmetrization map multiplies all symmetric tensors by $d!$, which annihilates them, and that strikes me as very funny behavior for a map called symmetrization!

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    $\begingroup$ In a ring, $d!$ may be non-invertible without being zero. It can even be a zero divisor without symmetrization's annihilating symmetric tensors (consider $((\mathbb{Z}/4\mathbb{Z})\oplus(\mathbb{Z}/4\mathbb{Z}))^{\otimes2}$). $\endgroup$
    – Alexey
    Jun 6, 2014 at 16:59
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    $\begingroup$ @Alexey: that's true. When I said "characteristic $p$" I had in mind specifically a field of characteristic $p$. Working over an arbitrary commutative ring only makes the corresponding representation theory even more complicated. $\endgroup$ Jun 6, 2014 at 17:33
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    $\begingroup$ Thanks for the explanations anyway, i will need to think about them. But i am mostly interested in modules over rings. $\endgroup$
    – Alexey
    Jun 6, 2014 at 17:34
  • $\begingroup$ At the end: "If you try to symmetrize without dividing by $d!$, then your symmetrization map multiplies all symmetric tensors by $d!$, which annihilates them, and that strikes me as very funny behavior for a map called symmetrization!" -- can you suggest a better symmetrization in this setting? :) $\endgroup$
    – Alexey
    Apr 2, 2015 at 15:43
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    $\begingroup$ Maybe you are right in that "symmetrization" is not an appropriate term for what i am looking for, because usually symmetrization is expected to leave symmetric things fixed. $\endgroup$
    – Alexey
    Apr 2, 2015 at 17:59
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Here is my partial answer or conjecture.

Let us look for a general notion of "symmetrization" (bad term?) of a function from a set to an Abelian semigroup in the following sense.

Given a finite group $G$ with a subgroup $H$, a set $X$ with a right $G$-action (a $G$-set), an Abelian semigroup $M$ with a right $G$-action by automorphisms (an Abelian $G$-semigroup), and an $H$-equivariant function $f\colon X\to M$, the $(G,H)$-symmetrization (bad term?) of $f$ must be a $G$-equivariant function $Q_{G,H}(f)\colon X\to M$. In addition, for a $G$-equivariant $f$, the equality $Q_{G,G}(f) = f$ must hold.

I have an impression that in some sense the only "reasonable" definition of such a functor $Q$ must be $$ Q_{G,H}(f)(x) =\sum_{g\in T}(f(x^{g^{-1}}))^g, \quad\text{where}\quad G =\bigsqcup_{g\in T}Hg $$ ($T$ is a right transversal for $H$ in $G$).

However, the formula in the spirit of Qiaochu Yuan's answer $$ P_{G,H}(f)(x) =\frac{\sum_{g\in T}(f(x^{g^{-1}}))^g}{|G:H|}, \quad\text{where}\quad G =\bigsqcup_{g\in T}Hg $$ is the good one in the case when the Abelian semigroup $M$ is replaced, for example, with a $\mathbb{Q}$-affine or a $\mathbb{Q}$-convex space $N$.

Terminology note

Probably "symmetrization" is not a good term for this because it is usually expected that "symmetrizations" be idempotent: symmetrizing symmetric things should not modify them. Maybe something like "folding" would be better?

Analogy with the adjugate matrix

Here is some intuitive analogy to explain why $Q_{G,H}$ can be considered to be "better" than $P_{G,H}$, despite failing to be idempotent. (It is more or less clear in which sense $P_{G,H}$ can be considered to be "better" than $Q_{G,H}$: $P_{G,H}$ is idempotent.)

The operation of taking the inverse matrix (over a field or a commutative ring) is a nice involution with good properties, but it is not always defined. Taking the adjugate matrix is not an involution, but it is always defined over a commutative ring, so it can be considered to be "better."


I may extend this answer if i manage to state the conjecture more precisely or justify the word "reasonable" better.

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  • $\begingroup$ As I've mentioned already, over a field of characteristic dividing the order of $G$ this not only fails to leave symmetric things fixed, it annihilates them! $\endgroup$ Apr 2, 2015 at 18:02
  • $\begingroup$ Thank you for making me think of this. But this only means that "symmetrization" is not a good word for this functor. I still would like to understand if it has any exceptional properties (maybe universal properties?). $\endgroup$
    – Alexey
    Apr 2, 2015 at 18:06

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